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Bulletin of the Iranian Mathematical Society Vol. 37 No. 1 (2011), pp 131-141.

A NOTE ON THE COMPARISON BETWEEN LAPLACE AND SUMUDU TRANSFORMS A. KILIC ¸ MAN*, H. ELTAYEB AND K. A. M. ATAN

Communicated by Mohammad Asadzadeh Abstract. In this paper, we discuss the existence of double Sumudu transform and study relationships between Laplace and Sumudu transforms. Further, we apply two transforms to solve linear ordinary differential equations with constant coefficients and non constant coefficients.

1. Introduction In the literature, there are several works on the theory and applications of integral transforms such as Laplace, Fourier, Mellin, Hankel, to name a few, but very little work on the power series transformation such as Sumudu transform. This is probably, because it is not widely used yet. The Sumudu transform was recently proposed by Watugala; see [12], [13]. The properties were established in [1] and subsequently applied to partial differential equations; see [2], [4], [7] and [9]. In our study, we use the following convolution notation: double convolution between two continuous functions F (x, y) and G(x, y) given by Z yZ x (1.1) F1 (x, y) ∗ ∗F2 (x, y) = F1 (x − θ1 , y − θ2 )F2 (θ1 , θ2 )dθ1 dθ2 ; 0

0

MSC(2010): Primary 35G15; Secondary 44A85, 44A35. Keywords: Laplace transform, Sumudu transform, convolution. Received: 22 July 2009, Accepted: 15 December 2009. ∗Corresponding author c 2011 Iranian Mathematical Society.

131

132

Kılı¸cman, Eltayeb and Atan

for further details and properties of the double convolutions and derivatives, we refer to [8] and [10]. The single Sumudu transform is defined over the set of the functions n o A = f (t) : ∃M, τ1 , τ2 > 0, |f (t)| < M e t/τj , if t ∈ (−1)j × [0, ∞) by 1 F (u) = S [f (t); u] = u

∞

Z

e−t/u f (t) dt,

u ∈ (−τ1 , τ2 ).

0

A sufficient condition for the existence of the Sumudu transform of a function f is of exponential order, that is, there exist real constants t + x M > 0, K1 , and K2 , such that |f (t, x)| ≤ M e K1 K2 . 2. Existence of the Sumudu Transform Theorem 2.1. If f is of exponential order, then its Sumudu transform S [f (t, x)] = F (v, u) exists and is given by Z ∞Z ∞ t x F (v, u) = e− v − u f (t, x)dtdx, where, points

1 1 i 1 u = η + τ and v 1 1 1 1 i u+v = η +µ+τ

1 η

0 i ξ.

= + The defining integral for F exists at i + + ξ in the right half plane η1 + µ1 > K11 + K12 .

and v1 = µ1 + ξi , we can express F (v, u) as: ∞Z ∞ t x − ηt − µx F (v, u) = f (t, x) cos + e dtdx τ ζ 0 0 Z ∞Z ∞ t x − ηt − µx −i f (t, x) sin + e dtdx. τ ζ 0 0

Proof. Using

1 u

0 1 µ

=

Then, for values of Z

+ Z

1 η

i τ

+

1 µ

∞Z ∞

0

>

≤ M 0

+

1 K2 ,

we have

t x − ηt − µx dtdx |f (t, x)| cos + e τ ζ

0 ∞Z ∞

Z

≤ M

1 K1

0

ηK1 η − K1

1 − η1 K1

e

1 t+ K1 − µ x

ηK2 µ − K2

2

dtdx

A note on the comparison between Laplace and Sumudu transforms

133

and Z

∞Z ∞

0

x − ηt − µx t dtdx |f (t, x)| sin + e τ ζ

0 ∞Z ∞

Z ≤ M 0

≤ M

0

ηK1 η − K1

1 − η1 K1

e

1 t+ K1 − µ x

ηK2 µ − K2

2

dtdx

which imply that the integrals defining the real and imaginary parts of F exist for value of Re( u1 + µ1 ) > K11 + K12 , and this completes the proof. Thus, we note that for a function f , the sufficient conditions for the existence of the Sumudu transform are to be piecewise continuous and of exponential order. We also note that the double Sumudu transform of function f (t, x) is defined in [5], by Z ∞Z ∞ x t 1 (2.1) F (v, u) = S2 [f (t, x); (v, u)] = e−( v + u ) f (t, x)dtdx uv 0 0 where, S2 indicates double Sumudu transform and f (t, x) is a function which can be expressed as a convergent infinite series. Now, it is well known that the derivative of convolution for two functions f and g is given by d d d (f ∗ g)(x) = f (x) ∗ g(x) or f (x) ∗ g(x) dx dx dx and it can be easily proved that Sumudu transform is: d d S dx (f ∗ g)(x); v = uS dx f (x); u S [g(x); u] d or uS [f (x); u] S dx g(x); u . The double Sumudu and double Laplace transforms have strong relationships that may be expressed either as (I) uvF (u, v) = £2 f (x, y); u1 , v1 or (II) psF (p, s) = £2 f (x, y); p1 , 1s where £2 represents the operation of double Laplace transform. In particular, the double Sumudu and double Laplace transforms interchange the image of sin(x + t) and cos(x + t). It turns out that

134


S2 [sin(x + t)] = £2 [cos(x + t)] = and

S2 [cos(x + t)] = £2 [sin(x + t)] =

u+v (1+u)2 (1+v)2 1 . (1+u)2 (1+v)2

3. Convolution Theorem In the next theorem, we prove double Sumudu transform of the double convolution. Theorem 3.1. Let f (t, x) and g(t, x) have double Sumudu transform. Then, double Sumudu transform of the double convolution of f and g, Z tZ x (f ∗ ∗g)(t, x) = f (ζ, η)g(t − ζ, x − η)dζdη, 0

0

exists and is given by (3.1)

S2 [(f ∗ ∗g)(t, x); v, u] = uvF (v, u)G(v, u).

Proof. By using the definition of double Sumudu transform and double convolution, we have Z ∞Z ∞ t x 1 S2 [(f ∗ ∗g)(t, x); v, u] = e−( v + u ) (f ∗ ∗g)(t, x)dtdx uv 0 Z t Z x0 Z ∞Z ∞ x t 1 −( v + u ) = e f (ζ, η)g(t − ζ, x − η)dζdη dtdx. uv 0 0 0 0 Let α = t − ζ and β = x − η, and using the valid extension of upper bound of integrals to t → ∞ and x → ∞, we have S2 [(f ∗ ∗g)(t, x); v, u] =

1 uv

∞

Z 0

Z

∞

ζ

η

e−( v − u ) dζdη

∞

Z

−ζ

0

∞

Z

α

β

e−( v − u ) g(α, β)dαdβ.

−η

Since both functions f (t, x) and g(t, x) have are zero, for t < 0, and x < 0, it follows with respect to lower limit of integrations that S2 [(f ∗ ∗g)(t, x); v, u] =

1 uv

∞

Z 0

Z 0

∞

ζ

η

e−( v − u ) dζdη

∞

Z 0

∞

Z

α

β

e−( v − u ) g(α, β)dαdβ.

0

Then, it is easy to see that S2 [(f ∗ ∗g)(t, x); v, u] = uvF (v, u)G(v, u). Thus, the relation between double Sumudu of convolution and double Laplace transform of convolution is given by 1 S2 [(f ∗ ∗g)(t, x); v, u] = £2 [(f ∗ ∗g)(t, x))] ; uv


135

see [11]. In fact, one can easily prove the following implications for existence: S2 (f (t)) ⇒ £2 (f (t)). Note that the inverse of implication need not be true; see [9]. n X ak

, where, n ≥ 0 and an 6= 0. Then, we define xk k=0 MP (x) to be a (1 × n) matrix in the following matrix product:   a1 a2 . . . an  a2 a3 . . an 0    1 1 1 1 a3 . . an 0 0  (3.2) MP (x) = ... n−1   . 2 3 x x x x  . . . . . .  an 0 . . . 0

Now, we let P (x) =

Thus, MP (x) defines a linear mapping of Cn into C in an obvious way. We shall write vectors y in Cn as row vectors or column vectors interchangeably, whichever is convenient, although when MP (x)y is to be computed and the matrix representation by Eq(3.2) of MP (x) is used, then, of course, y is written as a column vector: (3.3)

MP (x)y =

n n−i X 1 X ai+k yk , xi i=1

k=0

Cn .

for any y = (y0 , y1 , ..., yn−1 ) ∈ If n = 0, MP (x) defines a unique 0 linear mapping of {0} = C into C (empty matrix). In general, if n > 0 and f is n − 1 times differentiable on an interval (a, b) , with a < b, we shall write ϕ(f ; a; n) = f (a+), f 0 (a+), ..., f (n−1) (a+) ∈ Cn and φ(f ; b; n) = f (b−), f 0 (b−), ..., f (n−1) (b−) ∈ Cn . If a = 0, we write ϕ(f ; n) for ϕ(f ; 0; n). If n = 0, we define ϕ(f ; a; 0) = φ(f ; a; 0) = 0 ∈ C0 . In the next section, we give the Sumudu transform of higher derivatives.

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4. Sumudu Transform of Higher Derivatives Theorem 4.1. Let f be n times differentiable on (0, ∞) and let f (t) = 0, for t < 0. Suppose that f (n) ∈ Lloc. Then, f (k) ∈ Lloc , for 0 ≤ k ≤ n − 1, dom(Sf ) ⊂ dom(Sf (n) ), and for any polynomial P of degree n, (4.1)

P (u)S(y)(u) = S (f ) (u) + MP (u)ϕ(y, n),

for u ∈ dom(Sf ). Thus,

(4.2)

(Sf

(n)

1 )(u) = n (Sf )(u) − u

1 1 1 , n−1 , ..., n u u u

ϕ(f ; n).

In particular, if n = 2, then we have

(4.3)

(Sf 00 )(u) =

1 1 1 (Sf )(u) − 2 f (0+) − f 0 (0+). 2 u u u

Proof. We use induction on n. The result is trivially true if the case n = 0 and n = 1. Suppose now the result is true for some n ≥ 1 and let n+1 X ak P (x) = have degree n+1. Then, the first two statements follow by xk k=0 putting z = f 0 and using the induction hypothesis. Now, write P (x) = n · · P ak+1 a0 + x1 W (x), where, W (x) = . Then, P (D)f = a0 f + W (D)z xk k=0

and therefore, h · i h i · S P (D)f (u) = a0 S [f ] (u) + S W (D)z (u) − MW (u)ϕ(z; n) 1 1 = a0 S [f ] (u) + W (u) S [f ] (u) − f (0+) u u (4.4)

−

n−i n X 1 X ai+k+1 f ui i=1

k=0

(k+1)

(0+),


137

using Eq. (3.3) and z (k) = f (k+1) . The summation above can be written in the form: n−i+1 n X 1 X ai+k f ui i=1

(k)

(0+)

=

k=1

n−i+1 n X 1 X ai+k f ui i=1

(k)

(0+) −

k=0

=

n+1 X i=1

1 − u =

1 ui "

n−i+1 X

ai+k f

(k)

n X 1 a f (0+) i i u i=1

(0+)

k=0

# n X 1 1 an+1 f (0+) + ai f (0+) un ui−1 i=1

MP (u)ϕ(f ; n) −

1 W (u)f (0+). u

Thus, h · i 1 S P (D)f (u) = a0 + W (u) S [f ] (u) u 1 1 − W (u)f (0+) − MP (u)ϕ(f ; n) + W (u)f (0+) u u = P (u)S (f ) (u) − MP (u)ϕ(f ; n). In general, if f be differentiable on (a, b) with a < b, and f (t) = 0, for t < a or t > b and f (n) ∈ Lloc , then for all u, (4.5)

h · i h a i b S P (D)f (u) = P (u)(Sf )(u) − MP (u) e− u ϕ(f ; a; n) − e− u φ(f ; b; n) .

In particular if we consider y(t) = sin(t). Clearly, y 00 + y = 0, and so if we put f = yH, then f 00 + f = 0. Thus, D2 + 1 f = 0. Since dom(Sf ) is contained in (0, ∞) , we have from Eq. (4.1), and Eq. (4.2), with n = 2 and P (x) = x2 + 1, for u > 0, 1 1 1 0 1 0 0= + 1 S(f ) − . 1 0 1 u2 u u2 Since ϕ(y, 2) = (f (0), f 0 (0)) = (0, 1). Thus, from the above equation, we get u S [sin(t)H(t)] = 2 . u +1

138


Now let y be n times differentiable on (0, ∞) , zero on (−∞, 0) and satisfy ·

(4.6)

P (D)y =

n X

fi ∗ gi ,

i=1

under the initial conditions: y(0) = y0 , y 0 (0) = y1 , ..., y (n−1) (0) = yn−1 . Then, y (k) is locally integrable and Sumudu transformable for 0 ≤ k ≤ n, and thus Sumudu transform of Eq. (4.6) is given by  (4.7)

MP (u)ϕ(y, n) =

 1 1 1   ... u u2 un  

a1 a2 a3 . an

a2 a3 . . 0

. . . . .

. . an . .

. an 0 . .

an 0 0 . 0





      



y0 y1 . . .

   ,   

yn−1

and the non homogeneous term is a single convolution. In particular, if n = 2, we have (4.8) a a1 1 1 a1 2 + + a S(y)(u) = S (f ∗ g + f ∗ g ) (u) + 0 1 1 2 2 a2 u2 u u u2

a2 0

y0 y1

.

On taking the inverse Sumudu transform for Eq(4.8), we have the solution of Eq. (4.6) as follows: (4.9) y(t) = S −1

ha y u + a y + a y ui u [S (f1 ) S (g1 ) + S (f2 ) S (g2 )] 1 0 2 0 2 1 + S −1 , P (u) u2

provided that the inverse transform exists for each term in the right hand side of Eq. (4.9). Now, let us multiply the right hand side of Eq. n P (4.6) by a polynomial Ψ(t) = tk . Then, we obtain an equation with k=0

non constant coefficients in the form of (4.10)

h

·

i

Ψ(t) ∗ P (D)y =

n X i=1

fi ∗ gi ,


139

under the same initial conditions as given above. By taking the Sumudu transform for Eq. (4.10) and using the initial conditions, we have S [y] (u)

=

F (u)G(u) k!uk P (u) 

+

1 P (u)

1 1 1 ... u u2 un+1

    

a1 a2 a3 . an

a2 a3 . . 0

. . . . .

. . an . .

. an 0 . .

an 0 0 . 0





      

y0 y1 . . .

    .   

yn−1

By taking inverse of the Sumudu transform, we have the solution provided that the inverse exist. In particular, consider the differential equation in the form of y 000 − y 00 + 4y 0 − 4y = 2 cos(2t) − sin(2t) y(0) = 1, y 0 (0) = 4, y 00 (0) = 1.

(4.11)

On using Eq. (4.1), we have (4.12)

1 1 4 − + − 4 S(y)(u) = S [cos 2t ] + S [sin(2t)] + MP (u)ϕ(y, 4) u3 u2 u

and MP (u)ϕ(y, 4) = =

1 1 1 u u2 u3



  4 −1 1 1  −1 1 0   4  1 0 0 1

1 3 1 + 2 + 3. u u u

Then, we obtain: (4.13)

Y (u) =

u3 (2u + 1) (u2 + 3u + 1) + . (4u2 + 1)(1 − u + 4u2 − 4u3 ) (1 − u + 4u2 − 4u3 )

Replacing the complex variable u by 1s , Eq. (4.13) turns to: 1 s(s + 2) s(s2 + 3s + 1) (4.14) Y = 2 + . s (s + 4)(s2 + 4)(s − 1) (s2 + 4)(s − 1) Now in order to obtain the inverse Sumudu transform for Eq.(4.14), we use # " Z γ+i∞ 1 X Y ( ) 1 1 ds S −1 (Y (s)) = est Y = residues est s . s 2πi γ−i∞ s s

140


Thus, the solution of Eq. (4.11) is given by: y(t) =

13 1 sin(2t) − t cos(2t) + et . 8 4

Now, if we consider to multiply the left hand side of Eq. (4.11) by the non constant coefficient t2 , then Eq. (4.11) becomes (4.15) t2 ∗ y 000 − y 00 + 4y 0 − 4y = 2 cos(2t) − sin(2t) y(0) = 1, y 0 (0) = 4, y 00 (0) = 1. By applying a similar method, we obtain the solution of Eq. (4.15) in the form: 3 y1 (t) = cos(2t) − t sin(2t) + sin(2t). 2 Remark 4.2. A very interesting fact about the Sumudu transform is that the original function and its Sumudu transform have the same Taylor coefficients except for the factor n!. Thus, if f (t) = F (u) =

∞ X

∞ X

an tn ,

then

n=0

n!an tn ;

see [6]. Furthermore, Laplace and Sumudu transforms

n=0

of the Dirac delta function and the Heaviside function satisfy: S2 [H(x, t)] = £2 [δ(x, t)] = 1, 1 and S2 [δ(x, t)] = £2 [H(x, t)] = uv ; for details, see [3], where the concept of the Sumudu transform was generalized to distributions.

By using discontinuous or impulsive forcing terms, the Sumudu transform can be used to solve ODE(s) as well as PDE(s) in engineering problems. Acknowledgements The authors gratefully acknowledge that this research was partially supported by University Putra Malaysia under the Research University Grant Scheme 05-01-09-0720RU and Fundamental Research Grant Scheme 01-11-09-723FR. The authors also thank the referee for very constructive comments and suggestions.


141

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